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The continue statement
The continue statement causes the next iteration of the enclosing loop to start. When this is encountered in the loop , the rest of the statements in the loop are leave out and control passes to the condition.
Let us see an example that accepts a variable amount of numbers from the keyboard and prints the sum of only positive numbers.
e.g.
void main()
{
int num, total = 0;
do
cout << " enter 0 to quit ";
cin >> num; // equivalent to scanf()
if(num == 0)
break;
if(num < 0)
continue;
total+=num;
}
while(1);
cout << total;
In POS system, cashier computer connect to printer using serial rs232 usb or cable. We need to capture data from that connection, and parsing the data to get total amount for every
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programme for time table in c++
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if one won't explicitly call the destructor of a local; then how does he handle the above situation?
Describe problem with Runtime type identification? A: The run time kind identification comes at cost of performance penalty. Compiler maintains class.
Problems With Multiple Inheritance The following example presents a problem with multiple inheritance. class Aclass { public : void put()
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